Cohomology of a quotient of $S^3$

Question

Let $S^3 \subset \mathbb C^2$ be the unit sphere. We define an action of $S^1$ on $S^3$ by $\lambda \cdot (z_1,z_2) = (\lambda z_1,z_2)$ for $\lambda \in S^1$. What is the cohomology of the quotient space? I can not see what the quotient is (I know it cannot be a manifold)

Answer 1

Consider the map $ \pi \colon S^3\to D^2$ given by the projection on the second factor. Two elements $(z_1,z_2)$ and $(z_3,z_4)$ are sent to the same element if and only if $z_2=z_4$ which implies that $z_1$ and $z_3$ have the same norm. Assume $z_3\neq 0$, then the element $\frac{z_1}{z_3}$ has norm 1, i.e. it's an element of $S^1$. From here you should be able to prove that this map factor through the quotient and it defines is the required homeomorphism